Question

The population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.05 inches and a standard deviation of 0.2 inches. What is the probability that a sheet selected at random from the population is between 29.75 and 30.5 inches long?

Answer 1

92.10% probability that a sheet selected at random from the population is between 29.75 and 30.5 inches long.

Explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

`\mu = 30.05, \sigma = 0.2`

What is the probability that a sheet selected at random from the population is between 29.75 and 30.5 inches long?

This is the pvalue of Z when X = 30.5 subtracted by the pvalue of Z when X = 29.75

X = 30.5

`Z = (X - \mu)/(\sigma)`

`Z = (30.5 - 30.05)/(0.2)`

`Z = 2.25`

`Z = 2.25` has a pvalue of 0.9878

X = 29.75

`Z = (X - \mu)/(\sigma)`

`Z = (29.75 - 30.05)/(0.2)`

`Z = -1.5`

`Z = -1.5` has a pvalue of 0.0668

So there is a 0.9878 - 0.0668 = 0.9210 = 92.10% probability that a sheet selected at random from the population is between 29.75 and 30.5 inches long.